$\int \dfrac{(\ln{x})^2}{x}\,dx\,=$ Choose 1 answer: Choose 1 answer: (Choice A) A $ (\ln x)^3+C$ (Choice B) B $ (\ln x)^2+C$ (Choice C) C $ 2\ln x+C$ (Choice D) D $ \dfrac{(\ln x)^3}{3}+C$
Notice that we can rewrite the integral as $ \int (\ln x)^2\cdot \,\dfrac1x\, dx\,$. If we let $ {u=\ln(x)}$, then $du=\dfrac1x \, dx}$. Substituting gives us: $ \int ({\ln x})^2\cdot \,\dfrac1x\, dx}\,= \int u^2\, du}\,$ We recognize this antiderivative. $\begin{aligned}\phantom{\int\dfrac{e^x}{1+e^{2x}}dx}&= \int u^2\, {du}\,\\\\\\ &=\dfrac{u^3}{3}+C\end{aligned}$ We can now substitute back to find the antiderivative in terms of $x$. ∫ e x 1 + e 2 x d x = u 3 3 + C = ( ln x ) 3 3 + C \begin{aligned}\phantom{\int\dfrac{e^x}{1+e^{2x}}dx~}&=~\dfrac{u^3}{3}+C\\\\\\\ &=~\dfrac{(\ln x)^3}{3}+C\end{aligned} The answer: $\int \dfrac{(\ln{x})^2}{x}\,dx\,=~\dfrac{(\ln x)^3}{3}+C$